When ' x′×10−2mL methanol (molar mass =32g; density =0.792g/cm3 ) is added to 100 mL water (density =1g/cm3 ), the following diagram is obtained.
x= ............. (nearest integer) [Given: Molal freezing point depression constant of water at 273.15K∘ is 1.86Kkgmol−1 ]
Answer: 543
Solution
Depression in freezing point ΔTf=273.15−270.65=2.5 K. From ΔTf=Kfm: m=1.862.5=1.344 mol/kg. With 100 mL (=0.1 kg) water, moles of methanol =1.344×0.1=0.1344 mol, mass =0.1344×32=4.30 g, volume =0.7924.30=5.43 mL =543×10−2 mL, so x=543.